What is the center x^2+y^2+2x+4y=11 ?

The center x^2+y^2+2x+4y=11 is (-1,-2)

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center x^2+y^2+2x+4y=11

center

x^{2} + y^{2} + 2 x + 4 y = 11

(- 1, - 2)

Solution steps

x^{2} + y^{2} + 2 x + 4 y = 11

Rewrite

x^{2} + y^{2} + 2 x + 4 y = 11

in the form of the standard circle equation

(x - (- 1))^{2} + (y - (- 2))^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (- 1, - 2), r = 4

And the center is:

(- 1, - 2)

x^{2} + y^{2} + 2 y - 16 = 0

2 x^{2} + 2 y^{2} + 8 x + 7 = 0

(x - 2)^{2} + (y - 3)^{2} = 58

(x - 2)^{2} + y^{2} = 6

x^{2} + y^{2} = 144